Case statement fallthrough?

Question

In popular imperative languages, switch statements generally "fall through" to the next level once a case statement has been matched.

Example:

int a = 2; switch(a) {    case 1:       print "quick ";    case 2:        print "brown ";    case 3:        print "fox ";       break;    case 4:       print "jumped "; } 

would print "brown fox".

However the same code in bash

A=2 case $A in 2)   echo "QUICK"   ;& 2)   echo "BROWN"   ;& 3)   echo "FOX"   ;& 4)   echo "JUMPED"   ;& esac 

only prints "BROWN"

How do I make the case statement in bash "fall through" to the remaining conditions like the first example?

(edit: Bash version 3.2.25, the ;& statement (from wiki) results in a syntax error)

running:

test.sh:

#!/bin/bash A=2 case $A in 1)   echo "QUICK"   ;& 2)   echo "BROWN"   ;& 3)   echo "FOX"   ;& esac 

Gives:

./test.sh: line 6: syntax error near unexpected token ;' ./test.sh:
line 6: ;&'

Answer 1

The ;& and ;;& operators were introduced in bash 4.0, so if you want to stick with a five year old version of bash, you'll either have to repeat code, or use ifs.

if (( a == 1)); then echo quick; fi if (( a > 0 && a <= 2)); then echo brown; fi  if (( a > 0 && a <= 3)); then echo fox; fi if (( a == 4)); then echo jumped; fi 

or find some other way to achieve the actual goal.

(On a side note, don't use all uppercase variable names. You risk overwriting special shell variables or environment variables.)

Answer 2

Try this:

case $VAR in normal)     echo "This doesn't do fallthrough"     ;; fallthrough)     echo -n "This does "     ;& somethingelse)     echo "fall-through"     ;; esac